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On line Solution of Monovariate Equation:
    Input any unary equation directly, and then click the "Next" button to obtain the solution of the equation.
    It supports equations that contain mathematical functions.
    Current location:Equations > Monovariate Equation > The history of univariate equation calculation > Answer
    Overview: 6 questions will be solved this time.Among them
           ☆6 equations

[ 1/6 Equation]
    Work: Find the solution of equation 5x×3+3x×3-41 = 103 .
    Question type: Equation
    Solution:Original question:
     5 x × 3 + 3 x × 341 = 103
     Left side of the equation = 15 x + 9 x 41
                                             = 24 x 41
    The equation is transformed into :
     24 x 41 = 103

    Transposition :
     24 x = 103 + 41

    Combine the items on the right of the equation:
     24 x = 144

    The coefficient of the unknown number is reduced to 1 :
      x = 144 ÷ 24
        = 144 ×
1
24
        = 6 × 1

    We obtained :
      x = 6
    This is the solution of the equation.

[ 2/6 Equation]
    Work: Find the solution of equation 5x×3+25 = 457-3x×3 .
    Question type: Equation
    Solution:Original question:
     5 x × 3 + 25 = 4573 x × 3
     Left side of the equation = 15 x + 25
    The equation is transformed into :
     15 x + 25 = 4573 x × 3
     Right side of the equation = 4579 x
    The equation is transformed into :
     15 x + 25 = 4579 x

    Transposition :
     15 x + 9 x = 45725

    Combine the items on the left of the equation:
     24 x = 45725

    Combine the items on the right of the equation:
     24 x = 432

    The coefficient of the unknown number is reduced to 1 :
      x = 432 ÷ 24
        = 432 ×
1
24
        = 18 × 1

    We obtained :
      x = 18
    This is the solution of the equation.

[ 3/6 Equation]
    Work: Find the solution of equation 1.5(x-4) = 36 .
    Question type: Equation
    Solution:Original question:
     
3
2
( x 4) = 36
    Remove the bracket on the left of the equation:
     Left side of the equation =
3
2
x
3
2
× 4
                                             =
3
2
x 6
    The equation is transformed into :
     
3
2
x 6 = 36

    Transposition :
     
3
2
x = 36 + 6

    Combine the items on the right of the equation:
     
3
2
x = 42

    The coefficient of the unknown number is reduced to 1 :
      x = 42 ÷
3
2
        = 42 ×
2
3
        = 14 × 2

    We obtained :
      x = 28
    This is the solution of the equation.

[ 4/6 Equation]
    Work: Find the solution of equation (x+2)×2 = x+6.5 .
    Question type: Equation
    Solution:Original question:
     ( x + 2) × 2 = x +
13
2
    Remove the bracket on the left of the equation:
     Left side of the equation = x × 2 + 2 × 2
                                             = x × 2 + 4
    The equation is transformed into :
     2 x + 4 = x +
13
2

    Transposition :
     2 x x =
13
2
4
    i.e.
      x =
13
2
4

    Combine the items on the left of the equation:
      x =
13
2
4

    Combine the items on the right of the equation:
      x =
5
2
    This is the solution of the equation.
    This is the solution of the equation.

[ 5/6 Equation]
    Work: Find the solution of equation 4x+50 = 54+2x .
    Question type: Equation
    Solution:Original question:
     4 x + 50 = 54 + 2 x

    Transposition :
     4 x 2 x = 5450

    Combine the items on the left of the equation:
     2 x = 5450

    Combine the items on the right of the equation:
     2 x = 4

    The coefficient of the unknown number is reduced to 1 :
      x = 4 ÷ 2
        = 4 ×
1
2
        = 2 × 1

    We obtained :
      x = 2
    This is the solution of the equation.

[ 6/6 Equation]
    Work: Find the solution of equation 2x×3+2x-26 = 134 .
    Question type: Equation
    Solution:Original question:
     2 x × 3 + 2 x 26 = 134
     Left side of the equation = 6 x + 2 x 26
                                             = 8 x 26
    The equation is transformed into :
     8 x 26 = 134

    Transposition :
     8 x = 134 + 26

    Combine the items on the right of the equation:
     8 x = 160

    The coefficient of the unknown number is reduced to 1 :
      x = 160 ÷ 8
        = 160 ×
1
8
        = 20 × 1

    We obtained :
      x = 20
    This is the solution of the equation.



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